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**QUESTION** [NBHM(January)(2015) 1.1]

**Solve the following equation, given that it's roots are in arithmetic progression:**

[x^3-9x^2+28x-30 =0]

**DISCUSSION:** Now every root in the above mentioned equation is in arithmetic progression.

So let the roots be **a, a+d,** **a+2d**.

Now according to the cubic equation's root rule we get that

[a+(a+d)+(a+2d) = 9.......... eq(i)]

[a(a+d)+(a+d)(a+2d)+(a+2d)a = 28..........eq(ii)]

[a(a+d)(a+2d) = 30........... eq(iii)]

These above equations are gonna come in handy.

From eq (i) we clearly get [3a+3d = 9 & that means a+d =3]

So we will use this in our own ways. putting d = 3-a in our eq(iii) we get

[3a(6-a) = 30]

Which eventually will give us a quadratic equation like

[a^2-6a+10 = 0]

Calculate the roots by the same old same old and you have 3+i and 3-i.

Now you see that if **a is 3+i** and **a+d is 3** then clearly** a+2d will be 3-i** and similarly if you take **a = 3-i** then **a+2d is 3+i** and this is your required answer hope you liked it.

**QUESTION (1.2)**** Which of the statement is true?**

**a. Every group of order 51 is cyclic. b. Every group of order 151 is cyclic. c. Every group of order 505 is cyclic.**

**Discussion:** **To do this problem I am going to use a result given as a problem in I.N. Herstein (Section 2.9 problem 10, If it's difficult I will discuss the problem latter in another discussion) **

It is a simple enough problem if you know the result of the Herstein problem. It tells that if o(G) where G is the group is of the form pq where p and q are distinct primes and p>q.

Then **if q does not divide (p-1)** **then only** **G is cyclic**.

So 51 can be written as [3*17 = 51] and see that **3 does not divide 16 so 51 is clearly cyclic.**

Next up is 151.

This is a prime number guys and to prove that as a hint use the Fermat's Theorem which states that

[a^p \equiv a mod(p)] **Where p is a prime**.So it's obviously a cyclic group.

Last in the plate is 505 and now I give this to you to prove use the same result(theorem what you wish to call it) discussed in this section. For your relief the **answer is "no".**

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